A geostationary satellite is orbiting the earth at a height of 5R above that surface of the earth, R being the radius of the earth. The time period of another satellite in hours at a height of 2R from the surface of the earth is

5
10
6\(\sqrt{2}\)
\(\frac{6}{\sqrt{2}}\)

Solution

According to Kepler’s law of period T₂αR₃

\(\frac{T_{1}^{2}}{T_{2}^{2}}=\frac{R_{1}^{3}}{R_{2}^{3}}=\frac{(6R)^{3}}{(3R)^{3}}=8\)

\(\frac{24\times 24}{T_{2}^{2}}=8\)

\(T_{2}^{2}=\frac{24\times 24}{8}= 72 =36 \times 2\)

\(T_{2}=6\sqrt{2}\)

A planet moves around the sun. At a point P it is closest from the sun at a distance d₁ and has a speed v₁. At another point Q, when it is farthest from the sun at a distnace d₂ its speed will be

\(d_{1}^{2}v_{1}/d_{2}^{2}\)
\(d_{2}v_{1}/d_{1}\)
\(d_{1}v_{1}/d_{2}\)
\(d_{2}^{2}v_{1}/d_{1}^{2}\)

Solution

In planetary motion\(\vec{\tau }=0\Rightarrow \vec{L}\) constant

\(\vec{L}=\vec{r}\times \vec{p}(=m\vec{v})=mvr(∵\theta = 90º)\)

So m₁d₁v₁ = ₂d₂v₂(here r =d)

The weight of an object in the coal mine, sea level and at the top of the mountain,are respectively W₁, W₂ and W₃ then

W₁ < W₂ > W₃
W₁ = W₂ = W₃
W₁ < W₂ < W₃
W₁ > W₂ > W₃

Solution

At the surface of earth, the value of g =9.8m/sec². If we go towards the centre of earth or we go above the surface of earth, then in both the cases the value of g decreases.

Hence W₁=mg_mine, W₂=mg_{sea level}, W₃=mg_mounSo W₁ < W₂ > W₃(g at the sea level = g at the suface of earth)

Two planets of radii r₁ and r₂ are made from the same material. The ratio of the acceleration due to gravity g₁/g₂ at the surfaces of the two planets is

r₁/r₂
r₂/r₁
(r₁/r₂)²
(r₂/r₁)²

Solution

According to Gravitational Law

\(F=\frac{gM_{1}M_{2}}{r^{2}}\) .......(i)

Where M₁ is mass of planet & m₂ is the mass of anybody.Now according to Newton’s second law,body of mass m₂ feels gravitational acceleration g which is

F=M₂g ..........(ii)

So from(i) & (ii),we get g=\(\frac{GM_{1}}{r^{2}}\)

So the ratio of gravitational acceleration due to two planets is

\(\frac{g_{1}}{g_{2}}=\frac{M_{1}}{r_{1}^{2}}\times \frac{r_{2}^{2}}{M_{2}}=\frac{(4/3)\pi r_{1}^{3}\times \rho }{r_{1}^{2}}\times \frac{r_{2}^{2}}{(4/3)\pi r_{2}^{3}\times \rho }\)

\(\frac{g_{1}}{g_{2}}=\left ( \frac{r_{1}}{r_{2}} \right )\)
(both planet have same material, so density is same)

Taking the gravitational potential at a point infinte distance away as zero, the gravitational potential at a point A is –5 unit. If the gravitational potential at point infinite distance away is taken as + 10 units, the potential at point A is

– 5 unit
+ 5 unit
+ 10 unit
+ 15 unit

Solution

The gravitational potential V at a point distant ‘r’ from a body of mass m is equal to the amount of work done in moving a unit mass from infinity to that point.

\(V_{r}-V_{\alpha }=-\int_{\alpha }^{r}\vec{E}.d\vec{r}=-GM(1/r-1/)\alpha\)

\(=\frac{-GM}{r}\left ( As\: \vec{E}=\frac{-dv}{dr} \right )\)

(i)In the first case

when \(V_{\alpha }=0,V_{r}=\frac{-GM}{r}=-5\: unit\)

(ii)In the second case V_α= + 10 unit
V_r – 10 = – 5
or V_r= + 5 unit

A planet of mass 3 × 10²⁹ gm moves around a star with aconstant speed of 2 × 10⁶ ms^-1 in a circle of radii 1.5 × 10¹⁴ cm. The gravitational force exerte don the planet by the star is

6.67 × 10²² dyne
8 × 10²⁷ Newton
8 × 1026 N
6.67 × 10¹⁹ dyne

Solution

Gravitational force supplies centripetal force

\(∴F=\frac{mv^{2}}{r}=\frac{3\times 10^{29}\times (2\times 10^{8})^{2}}{1.5\times 10^{14}}\) dynes

= 8 × 10³¹ dyne = 8 × 10²⁶ N (∵ 1N = 10⁵ dyne.)

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